math mark scheme confusion

for the 6c in the tan(k) part.. the mark scheme says
k is now in the 4th quadrant, where cos k
is positive. So the value of cos k remains
the same and there is no change to tan k .
How though?? Shouldn't it tank be negative as it is now in the 4th quadrant??
solution to tank in 6bii was tank=(x/sqrt(1-x^2)) in case that's needed

If x is negative, then the bold is negative as x is negative. Im presuming no change means the expression does not change, but obviously the sign changes (like sin) between quadrants 1 and 4.

but tank does become negative right?

In quadrant 4, so k in -pi/2..0, then
arcsin(x) = k
and x is negative and
tan(k) = sin(k)/cos(k) = x / sqrt(1-x^2)
is negative